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Finding all the forces in this system

  1. Aug 1, 2017 #1
    1. The problem statement, all variables and given/known data
    System.png

    I am trying to identify every force in this system and prove that it is in mechanical equilibrium when tan(θ)=⅓. Initially I had to solve it by finding when the derivative of potential with respect to θ was 0, but now I am just trying to resolve the forces.
    2. Relevant equations
    Upwards forces = Downwards Forces
    Clockwise moments = Anticlockwise moments
    3. The attempt at a solution
    First I let m=3 and g=9.8. I worked out the tension T in the string knowing that it must equal the weight of the particle, giving 29.4N. I worked out the forces on each of the pulleys assuming the tension on either side of them is equal. At A it will be 2Tcos(θ/2) which is 58.0 and at D it will be 2Tcos(1/2*(90-θ)) for which I got 47.7. Also you know that the force on E by the string will be equal to the tension so 29.4N. A quick check of these forces balancing horizontally shows that I am right. Now where I am stuck is on the vertical forces. I am not sure how the reaction forces at B and C and the reaction force between the two rods at A play a role and how could I work them out to show that they balance?
     
    Last edited: Aug 1, 2017
  2. jcsd
  3. Aug 1, 2017 #2
    The image is not visible, so it is impossible to comment on your problem.
     
  4. Aug 1, 2017 #3
    I have edited it so the image now appears.
     
  5. Aug 1, 2017 #4
    Well, start by drawing a free body diagram for each leg, and do not forget to include horizontal forces as well.
     
  6. Aug 1, 2017 #5
    Since they are on a smooth horizontal surface, will there only be vertical forces on the bottom of the legs? And where the two rods meet at A, will there be both horizontal and vertical forces?
     
  7. Aug 1, 2017 #6
    Don't forget the point where the cord is secured.
     
  8. Aug 1, 2017 #7
    Forces Resolved.png
    I separated the rods to make the forces easier to label. Are these forces correct? I am not sure about the forces RAH and RAV. Are they in the right directions?
     
  9. Aug 1, 2017 #8
    I don't propose to work your problem. Instead, I suggest that you simply ask yourself to justify every force you have shown. If you can do so, you are probably good to go. If not, then maybe it is time to re-think.

    As to Rah and Rav, you can assume them in any direction that suits you. If you keep the signs straight, it will all work out in the end.

    Looks to me like the intent of the problem was to work with the mass = m, not necessarily 3 kg. I'd suggest you go back to expressing weights as m*g rather than using numerical values.
     
  10. Aug 1, 2017 #9

    haruspex

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    You can find the relationship between RB and RC easily by taking moments for the whole system.
    It take it that the RAV are the vertical forces the rods exert on each other. According to Newton, action and reaction are ..... ?
     
  11. Aug 2, 2017 #10
    Right, so one of the vertical reaction forces should be in the opposite direction, thanks. The force on the pulley at A, will that act on the right-hand rod as well? The original diagram makes it look like it only acts on the left one, however since it is at A should it act on both of them equally?
     
  12. Aug 2, 2017 #11

    haruspex

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    It is unlikely to act on them equally because that force is not vertical. I would take it as acting on the left hand rod only.
     
  13. Aug 2, 2017 #12
    I have taken moments about A on both rods and found that both reaction forces equal 73.5. This value seems to balance the vertical forces on both sides so I think it is right. The final question I have then is whether there is some intuitive reason why they have the same reaction force as it doesn't seem obvious.
     
  14. Aug 2, 2017 #13

    haruspex

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    You mean at B and C? You can treat the whole structure as one body with five external forces, gravity on the two rods and on the suspended mass, opposing the two ground reaction forces. The gravitational forces are symmetric about the centre line, so the reaction forces must be equal.
     
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